Question: Let $g$ be a differentiable function with $g(-1)=5$ and $g'(-1)=2$. What is the value of the approximation of $g(-0.9)$ using the function's local linear approximation at $x=-1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $5.1$ (Choice B) B $5.2$ (Choice C) C $5.3$ (Choice D) D $5.4$
Solution: The local linear approximation of $g$ at $x=-1$ is achieved using the equation of the line tangent to $g$ at $x=-1$. Let $L(x)$ represent this equation. We can find $L(x)$ using the general formula for the tangent to the graph of function $u$ at $x=a$ : $y=u'(a)(x-a)+u(a)$ [Is there a way to find this formula without memorizing?] In our case, $L(x)=g'(-1)(x+1)+g(-1)$. Plugging $g(-1)=5$ and $g'(-1)=2$, we obtain $L(x)=2(x+1)+5$. To approximate $g(-0.9)$, all we need is to plug $x=-0.9$ into $L(x)$. $\begin{aligned} L(-0.9)&=2(-0.9+1)+5 \\\\ &=2(0.1)+5 \\\\ &=5.2 \end{aligned}$ In conclusion, the approximation of $g(-0.9)$ using the function's local linear approximation at $x=-1$ is $5.2$.